We know that 2k = 9m = 7s = 15b. In other words: 2k, 9m, 7s and 15b each equal the same integer number, but what is that number? It would have to be cleanly divisible by 2, 9, 7, and 15. In order to minimize the number of packets owned by the group as a whole, we would want to find the smallest such number. That would be least common multiple of 2, 9, 7, and 15.

Note that we leave out a 3, compared to the list of all factors above. Remember that we only include the factors needed to build each of the starting numbers individually. With two 3’s and a 5, we could make either 9 or 15. That’s good enough for a least common multiple.

Now we know that 2k, 9m, 7s, and 15b each equals 630. Let’s find how many packets each person has.

(E) is another possible multiple of 2, 9, 7 and 15. (2)(9)(7)(15) = 1890. But it’s not the least common multiple. It’s the trap for those who don’t omit a redundant 3, and who also forget to finish the solving. (The value of 2k = 9m = 7s = 15b is not the answer, k + m + s + b is the answer.)

The correct answer is B.

Woo-hoo! But I actually think my own method, with the compound ratio, is quicker.